Question

A uniform disc of radius R lies in x-y plane with its center at origin. Its moment of intertia about the axis x=2R and y=0 is equal to the moment of intertia about the axis y=d and z=0, where d is equal to.

• A. $2R$
• B. $\frac{\sqrt{17}}{2}R$
• C. $\sqrt{13}R$
• D. $\frac{\sqrt{15}}{2}R$

Answer 1

Answer: (B)

$\frac{\sqrt{17}}{2}R$

$\begin{array}{c}\text{casel: moment of Inertia about axis perpendicular}\\ \text{to its plane and passing through its centre}\\ \text{is renown to be}=\frac{M{R}^2}{2}\\ \text{By Parallel axis theorem moment of Inertia}\\ \text{about given axis will be -}\\ I=\frac{M{R}^2}{2}+M{d}^{2}d=2R\\ \Rightarrow I=\frac{M{R}^2}{2}+4M{R}^2=\frac{9}{2}M{R}^2\end{array}$

$\begin{array}{c}\text{Case}2:\text{By Perpendicular axis theorem moment of}\\ \text{Inertia passing through its plane and cuntre}\\ \text{is}\to I_{11}+I_{11}=I_L=\frac{M{R}^2}{2}\\ \Rightarrow I_{11}=\frac{M{R}^2}{4}\\ \Rightarrow I_{AB}=\frac{M{R}^2}{4}\end{array}$

$\begin{array}{c}\text{Therefore by parallel axis theorem moment of}\\ \text{Inertia about given axis will be -}\\ I=\frac{M{R}^2}{4}+M{d}^2\\ \text{Now this is same to be moment of Inertia in Case}\\ \Rightarrow \frac{M{R}^2}{4}+M{d}^2=\frac{9}{2}M{R}^2\\ \Rightarrow d=\frac{\sqrt{17}}{2}R\\ \text{Henu, option (B) is correct.}\end{array}$